Test: Coordination Chemistry- 1


20 Questions MCQ Test Inorganic Chemistry | Test: Coordination Chemistry- 1


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QUESTION: 1

Which of the following represents a chelating ligand

Solution:

Because DMG is a polydentate ligand that binds the central atom nickel forming a ring-like structure.

QUESTION: 2

The complex compounds which result from the coordination of carbon monoxide are known as 

Solution:

Metal carbonyls are coordination complexes of transition metals with carbon monoxide ligands. Metal carbonyls are useful in organic synthesis and as catalysts or catalyst precursors in homogeneous catalysis,

Fe(CO)5,Ni(CO)4 etc. 

QUESTION: 3

Which of the following cannot act as a ligand?

Solution:
  • NHis a neutral ligand.
  • NOis a cationic ligand.
  • Cl– is an anionic ligand.
  • However, BFis not a ligand as it does not have lone pairs to donate and can not share a pair of electrons.

 

QUESTION: 4

As a ligand Cl is:

Solution:

Cl-  is π-donor (minus charge form σ bond with the metal orbitals, eg, t1u and a1g orbitals in Oh field. The lone pair of electron on Cl will try to form π-bonding with t2g orbitals of metal, electron cloud is transferred from Cl to metal d orbitals).

QUESTION: 5

In the complexes [Fe(CN)6]3– and [Co(en)3]3+, the coordination number of iron and cobalt are respectively (en = ethylenediamine):

Solution:
  • In 1st compound, CN is a unidentate ligand and hence it will surround the Fe atom from 6 different sides.
  • On the other hand, the 2nd compound contains bidentate ligand so each (en) will have two different sites of donating electrons and hence it required 3 (en) atoms for surrounding Co atom from 6 different sides.
    Hence A is correct.
QUESTION: 6

The oxidation number, coordination number and magnetic moment in the following complex, [Cr(C2O4)2 (NH3)2]- is:

Solution:
  • Oxidation number:
    x + (-2 x 2) + 0 x 2 = -1
    x = +3
  • The coordination number is 6 because C2O42-  is a bidentate ligand and NH3 is monodentate. 
  • Magnetic moment = √n(n+2)
    where n is no. of unpaired electrons.
    For Cr, n = 3 so the Magnetic moment will be √15 BM.
QUESTION: 7

Following are the transition metal ions of 3d series: Ti4+,V2+,Mn3+,Cr3+
(Atomic numbers: Ti=22,V=23,Mn=25,Cr=24)
Which ion is most stable in an aqueous solution and why?

Solution:

Ti4+ is most stable as it has completely filled 3p orbital.

QUESTION: 8

Octahedral complex of Cr (III) will be:

Solution:
QUESTION: 9

Which of the following is a spin paired complex ion?

Solution:
  • In [Co(NH3)6]3+, the oxidation state of cobalt is +3. Ammonia is a strong field ligand so it pairs up 4 unpaired electrons and frees up 2,3−d orbitals.
  • These 3−d orbitals are involved in hybridisation with one 4s and three 4p orbitals forming an inner orbital complex, so hybridisation of [Co(NH3)6]3+ is d2sp3.
QUESTION: 10

Compound ‘X’ has the molecular formula CrCl2Br.6H2O. It can show which type of isomerism:
(I) Hydrate isomerism
(II) Ionization isomerism
(III) Geometrical isomerism
(IV) Optical isomerism

Solution:
QUESTION: 11

What is the change in the oxidation state of cobalt in the following reaction?

[Co(NH3)4 Cl2]+ + H2O → [Co(NH3)4 (H2O)Cl]2+ + Cl

Solution:
QUESTION: 12

The first-row transition metal complexes having tetrahedral geometry are high–spin due to:

Solution:
QUESTION: 13

The zero magnetic moment of octahedral K2NiF6 is due to:

Solution:

Oxidation state of Ni in [NiF6]2- :
x − 6 = −2
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand.
⇒ low spin d6 Ni(IV) complex.
⇒ unpaired electrons = 0. 
⇒ zero magnetic moment.

QUESTION: 14

What is the spin only magnetic moment value in (Bohr Magneton units) of Cr(CO)6?

Solution:
  • The electron configuration is [Ar]3d54s1.
  • We have to accommodate the 6 Ligands and the fact that CO is a strong ligand, this results in d2sp3 hybridization.
  • Therefore, there are no unpaired electrons in Cr(CO)6. So n = 0. Thus, the spin only magnetic moment is also 0.
QUESTION: 15

The volume (in mL) of 0.1 M AgNO3 required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:

Solution:

QUESTION: 16

Which of the following elements do not form a complex with EDTA? 

Solution:

Be is the only group 2 element that does not form a stable complex with [EDTA]4−⋅Mg2+ and Ca2+ have the greatest tendency to form complexes with [EDTA]4−.

QUESTION: 17

When EDTA solution is added to Mg2+ ion solution, then which of the following statements is not true?

Solution:
  • When EDTA solution is added to the Mg2+ ion solution, all six-coordinate sites of Mg2+ are occupied, and a colourless [Mg−EDTA]2− chelate is formed. This reaction also results in the lowering of the pH of the solution.
  • So, the correct answer is 'Four coordinates sites of Mg2+ are occupied by EDTA and the remaining two sites are occupied by water molecules.'
QUESTION: 18

The total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is:

Solution:

The total number of geometrical isomers for the complex [RhCl(CO)(PPh3​)(NH3)] is since it is a square planar. 

QUESTION: 19

Which one the following octahedral complexes will be distorted:

Solution:
  • [Cr(H2O)6]2+ is distorted because of more number of electrons i.e. 4 electrons (electron-electron repulsion).
  • But in the case of Mn & Fe both of them has 3d5 configuration which is more stable & decreases the chances of distortion.
  • In the case of Cr3+ less number of electrons are present, so, less electron-electron repulsion hence lesser will be the distortion.
  • When there is unsymmetrical filling of electron in Eg orbital occur, the shape will distorted e.g. in the case of d4. This is known as Jahn-Teller distortion.
    Hence A is correct.
QUESTION: 20

The compound which exhibits Jahn-Teller distortion is:

Solution: